Graph tan(theta)<0 , sec(theta)<0

$\tan (θ) < 0$ , $\sec (θ) < 0$

Step 1

Solve for $θ$ .

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ < \arctan (0)$

Simplify the right side.

The exact value of $\arctan (0)$ is $0$ .

$θ < 0$

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + 0$

Add $π$ and $0$ .

$θ = π$

Find the period of $\tan (θ)$ .

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

The period of the $\tan (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = π n, π + π n$ , for any integer $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Find the domain of $\tan (θ)$ .

Set the argument in $\tan (θ)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$θ = \frac{π}{2} + π n$ , for any integer $n$

The domain is all values of $θ$ that make the expression defined.

${θ | θ \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Use each root to create test intervals.

$0 < θ < \frac{π}{2}$

$\frac{π}{2} < θ < π$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $0 < θ < \frac{π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < θ < \frac{π}{2}$ and see if this value makes the original inequality true.

$θ = 1$

Replace $θ$ with $1$ in the original inequality.

$\tan (1) < 0$

The left side $1.55740772$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $\frac{π}{2} < θ < π$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < θ < π$ and see if this value makes the original inequality true.

$θ = 2$

Replace $θ$ with $2$ in the original inequality.

$\tan (2) < 0$

The left side $- 2.18503986$ is less than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$0 < θ < \frac{π}{2}$ False

$\frac{π}{2} < θ < π$ True

$0 < θ < \frac{π}{2}$ False

$\frac{π}{2} < θ < π$ True

The solution consists of all of the true intervals.

$\frac{π}{2} + π n < θ < π + π n$ , for any integer $n$

Step 2

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 3

Plot each graph on the same coordinate system.

$\tan (θ) < 0$

$\sec (θ) < 0$

Step 4

Graph tan(theta)<0 , sec(theta)<0

Related Questions

Support LCMGCF.com by disabling your adblocker.